Calendar Count 1: Which date falls on which day?

1. Background

While watching the horror movie "Friday the 13th", one question would have definitely crossed your mind "Which date in real time is referred to as 'Friday the 13th'?".

Also, many of us have wondered with the questions like:

"On which day India got independence? Was it a Sunday, Monday or some other day?"

"On which day will I be celebrating my 50th birthday?"

"What were the last day in the BC era and first day in the AD era?"

Well, you can get to know the answers to these and many other similar questions on your own.

2. Gregorian and Julian Calendars

Many different calendars had been and have been used in various parts of the world. But there is only one which is accepted world-wide: "the Gregorian calendar" (commonly known the English calendar).

The Gregorian calendar was introduced by Pope Gregory XIII in October 1582. The European countries gradually accepted this calendar to Julian Calendar which was in use since 1 January 45 BC (709 AUC). Great Britain and the British Empire (including the USA) accepted the Gregorian Calendar in 1752. This resulted in skipping 11 days in the month of September (as in the figure below) that year.

Here I am going to give you some simple formulas to accurately determine the day of a specific date in Gregorian (since 1752) calendar and Julian (before 1752) calendar.

Julian calendar had a leap year once in four years.

In Gregorian calendar, leap year occurs once in four years except for the years that are divisible by 100. But these centurial years are leap years if they are exactly divisible by 400. For example, the years 1900 and 2100 are not leap years, but the year 2000 is.

3. The Procedure

Now, let us look into the simple method to find the day corresponding to a given date in few simple steps.

3.1. Since 14 September 1752

Step 1: Know today's day and date. Today is 17 June 2020 and it is a Wednesday.

Step 2: Find the day corresponding to the date in the current month by subtraction. Then, divide the result by 7 and add/subtract the remainder.

For example, if we wanted to find on which day the last republic day fell, then calculate the day corresponding to 26 of the current month. Among 26 (required date) and 17(today's date), 26 is higher. Subtracting 17 from 26, we get 9. Dividing 9 by 7 yields 2 as a remainder. Add 2 to today's day. Hence, 26 June 2020 will fall on Friday(two days after Wednesday).

If the required date was less than today's date, then subtract the remainder to today's day.

Step 3: Find the day corresponding to the required month and date in the current year. If you encounter the last day of a month add 2 or 3 depending on whether the month has 30 or 31 days. In case of February, add 1 if it is a leap year, else do not add anything. Divide the result by 7 and add/subtract to/from the result obtained in Step 2.

For example, in step 2 example, we found that 26 June 2020 would fall on Friday. Our aim is to find the day corresponding to 26 January of the current year. For that we need to pass through the last days of May, April, March, February (leap year) and January. Here there are 3 months with 31 days, 1 month with 30 days and February of a leap year. Adding these together,

3*3 + 1*2 + 1*1 = 12

Dividing 12 by 7, we get 5 as remainder. Subtracting 5 from Friday (day corresponding to required date of current month), we obtain that last republic day fell on a Sunday.

Step 4: If the required year is in the past, subtract the required year from the current year. If the required year is in the future, subtract the current year from the required year. Let the answer be 'x'. Divide 'x' by 4 and add the quotient to 'x'. Let the sum be 'y'. If in-between, you encounter 'n' centurial years not divisible by 400, subtract 'n' from 'y'. Let the result be 'z'.

If the required year is a leap year, then do not make any further changes. If it is in the past, is a non-leap year and the date falls after 28 February, add 1 to 'z'. If the required year is in the future, is a non-leap year and the date falls prior to 28 February, add 1 to 'z'. In all other cases, do not add anything. Let the obtained result be 'w'.

Divide 'w' by 7 and add/subtract the remainder to/from the result obtained in Step 3 depending on whether the date is in the future/past.

Example 1: 23 May 1932 (Past)

Solution:

Step 1: 17 June 2020 (Today) is a Wednesday.

Step 2:

23 - 17 = 6.

Remainder when dividing 6 by 7 = 6.

Add 6 to today's day obtaining 23 June 2020 as a Tuesday.

Step 3:

Encountered last day of May only which has 31 days.

Got 3 here.

Subtract 3 from the result obtained in Step 2 as we are going to the month which had already got over in the current year.

Hence, 23 May 2020 was a Saturday.

Step 4:

Required year is in the past.

x = 2020 - 1932 = 88

Quotient when x is divided by 4 is 22.

So, y = 88 + 22 = 110

Centurial year encountered is 2000. But, it is divisible by 400.

Thus, n = 0.

And, z = y - n = 110

The required year is a leap year. Hence, no further alteration. Hence, w = z = 110.

Now, the remainder obtained when w is divided by 7 = 5.

Hence, subtract 5 from the result obtained in Step 3.

Therefore, 23 May 1932 was a Monday.

Example 2: 2 February 2222 (Future)

Solution:

Step 1: 17 June 2020 (Today) is a Wednesday.

Step 2:

17 - 2 = 15.

Remainder when dividing 15 by 7 = 1.

Subtract 1 to today's day obtaining 2 June 2020 as a Tuesday.

Step 3:

Encountered last days of May, April, March and February.

Two of these months have 31 days, one has 30 days and the remaining one is a February of a leap year.

Adding the numbers here, 2*3 + 1*2 + 1*1 = 9.

Remainder when dividing 9 by 7 = 2.

Subtract 2 from the result obtained in Step 2 as we are going to the month which had already got over in the current year.

Hence, 2 February 2020 was a Sunday.

Step 4:

Required year is in the future.

x = 2222 - 2020 = 202

Quotient when x is divided by 4 is 50.

So, y = 202 + 50 = 252

Centurial years encountered are 2100 and 2200 both of which are not divisible by 400.

Thus, n = 2.

And, z = y - n = 252 - 2 = 250

The required year is in the future, is a leap year and falls prior to 28 February. Thus, add 1 to z. Hence, w = z + 1 = 250 + 1 = 251.

Now, the remainder obtained when w is divided by 7 = 6.

Hence, add 6 to the result obtained in Step 3.

Therefore, 2 February 2222 is a Saturday.

3.2. 3 September 1752 to 13 September 1752

These days do not exist in the history of Britain and the British Empire. People in these areas slept on 2 September 1752 and woke up on 14 September 1752.

3.3. Prior to 2 September 1752

Julian calendar was in use in these days. But in some of those years, the legal year started started on 25 March. Therefore, the year in official record may be one year more than the calculated one.

Steps 1 to 3: Same as in subsection 3.1.

Step 4: Subtract the required year from the current year. Let the answer be 'x'. Divide 'x' by 4 and add the quotient to 'x'. Let the sum be 'y'. Subtract 2+11 = 13 from 'y' (Two centurial years 1800 and 1900 were non-leap years and skipping 11 days in September 1752). Let the result be 'z', that is, z = y - 13.

If the required year is a leap year, then do not make any further changes. If it is a non-leap year and the date falls after 28 February, add 1 to 'z'. In all other cases, do not add anything. Let the obtained result be 'w'.

Divide 'w' by 7 and subtract the remainder from the result obtained in Step 3.

Example 3: 11 November 1111

Solution:

Step 1: 17 June 2020 (Today) is a Wednesday.

Step 2:

17 - 11 = 6.

Remainder when dividing 6 by 7 = 6.

Subtract 6 from today's day obtaining 11 June 2020 as a Thursday.

Step 3:

Encountered last days of June, July, August, September and October.

Three of these months have 31 days and two have 30 days.

Adding the numbers here, 3*3 + 2*2 = 13.

Remainder when dividing 13 by 7 = 6.

Add 6 to the result obtained in Step 2 as we are going to the month which has not yet come in the current year.

Hence, 11 November 2020 will be a Wednesday.

Step 4:

x = 2020 - 1111 = 909

Quotient when x is divided by 4 is 227.

So, y = 909 + 227 = 1136

And, z = y - 13 = 1123

The required year is a non-leap year and the date falls after 28 February. Thus, add 1 to z.

Hence, w = z + 1 = 1124

Now, the remainder obtained when w is divided by 7 = 4.

Hence, subtract 4 from the result obtained in Step 3.

Therefore, 11 November 1111 was a Saturday.

Note: The AD and BC system of naming a year came into effect somewhere in the sixth century AD. So, I assume AUC (Ab Urbe Condita) years were used before that in ancient Rome. Important conversion from AUC to BC/AD are:

AUC 1 = 753 BC

AUC 709 = 45 BC (Julian calendar came into effect)

AUC 753 = 1 BC

AUC 754 = AD 1

Keeping these in mind, let us compute the first day of the AD era and the last day of BC era.

Example 4: 1 January AUC 754 (AD 1)

Solution:

Step 1: 17 June 2020 (Today) is a Wednesday.

Step 2:

17 - 1 = 16.

Remainder when dividing 16 by 7 = 2.

Subtract 2 from today's day obtaining 1 June 2020 as a Monday.